题目

已知数列an中,a1=3,an+1=2an-1,设bn=2n/anan+1,求证：数列bn的前n项和Sn＜1/3

提问时间：2021-01-24

答案

数列an中,a1=3,a=2an-1,∴a-1=2(an-1),∴an-1=(a1-1)*2^(n-1)=2^n,∴an=2^n+1,∴bn=2n/{(2^n+1)(2^(n+1)+1]},∴Sn=2/(3*5)+4/(5*9)+……+2n/{(2^n+1)(2^(n+1)+1]}=1/3-1/5+1/5-1/9+……+[n/2^(n-1)][1/(2^n+1)-1/[2^...